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Chapter 4: QUADRATIC EQUATIONS

Grade 10 · Mathematics

Quadratic Equations

Learn how to identify, factorise, and solve equations of degree 2 — and see how they appear in everyday life!

📋 What Is It? ✂️ Factorisation 🔢 The Formula △ Discriminant

📋

1. What Is a Quadratic Equation?

A quadratic equation is any equation that can be written in the form:

ax² + bx + c = 0

where a, b, c are real numbers and a ≠ 0

The word quadratic comes from quadratus, Latin for “square” — because the variable x is squared. The highest power of x is 2, which is what makes it a quadratic equation rather than linear (degree 1) or cubic (degree 3).

Diagram labelling each part of the quadratic equation ax² + bx + c = 0

🔍 How to Tell if an Equation is Quadratic

✅ These ARE Quadratic

3x² + 5x − 2 = 0 degree 2 ✓
x² − 9 = 0 b=0, still quadratic ✓
5x² = 0 b=0, c=0, still quadratic ✓
x(x+4) = 7 expands to x²+4x−7=0 ✓

✗ These are NOT Quadratic

4x + 7 = 0 degree 1 (linear) ✗
x³ − x = 0 degree 3 (cubic) ✗
√x = 5 not a polynomial ✗
1/x + 3 = 0 has x in denominator ✗

💡 What Does “Root” Mean?

A root (or solution) is a value of x that makes the equation true — i.e., makes the left side equal zero. Every quadratic equation has at most 2 roots. The graph of a quadratic is a curved shape called a parabola, and the roots are the x-values where the parabola crosses the x-axis.

Parabola graphs: upward U-shape when a>0 and downward arch when a<0, showing roots where curve crosses x-axis

🏗️

2. Turning Real Problems into Equations

Many everyday situations can be described with a quadratic equation. The key skill is reading the problem carefully, choosing a variable, and building the equation step by step.

Original worked example: a garden whose length is 3 metres more than its width, with area 252 m², modelled as w²+3w−252=0

WORKED EXAMPLE 1

Garden Dimensions

A rectangular garden has an area of 252 m². Its length is 3 metres more than its width. Find the width and length.

Let width = w metres → length = (w + 3) metres
Area = length × width → (w + 3) · w = 252
w² + 3w = 252 → w² + 3w − 252 = 0

Now solve this quadratic (shown in the factorisation section below).

WORKED EXAMPLE 2

Football Ticket Sales

A sports club sells tickets for a match. When the price per ticket is ₹x, the number of tickets sold is (80 − x). The total revenue on a particular day was ₹1500. Find the ticket price.

Revenue = price × quantity → x(80 − x) = 1500
80x − x² = 1500 → x² − 80x + 1500 = 0

This is our quadratic. We solve it to find x = 30 or x = 50 (both valid prices).

WORKED EXAMPLE 3

Two Consecutive Numbers

The product of two consecutive even numbers is 168. Find the numbers.

Let the smaller number = n (even) → next even = n + 2
n(n + 2) = 168 → n² + 2n = 168 → n² + 2n − 168 = 0

Solving gives n = 12, so the two numbers are 12 and 14. Check: 12 × 14 = 168 ✓

✂️

3. Solving by Factorisation

Factorisation works by writing ax² + bx + c as a product of two smaller (linear) expressions, then setting each one to zero. The trick is splitting the middle term bx into two terms whose coefficients multiply to give a × c and add to give b.

4-step diagram showing how to factorise a quadratic equation by splitting the middle term

📌 The Middle-Term Split — Quick Recipe

Find two numbers p and q where:

p × q = a × c
p + q = b

Then rewrite and group:

Replace bx with px + qx
Group in pairs and factorise

📝

Factorisation — Worked Examples

EXAMPLE A

Solve: x² + 7x + 12 = 0

a=1, b=7, c=12 → a×c = 12 → Need p×q=12 and p+q=7
✔ p=3, q=4 (since 3×4=12 and 3+4=7)

x² + 7x + 12 = x² + 3x + 4x + 12
= x(x+3) + 4(x+3)
= (x+3)(x+4) = 0

x = −3

x = −4

Check: (−3)²+7(−3)+12 = 9−21+12 = 0 ✓

EXAMPLE B

Solve: 2x² − 7x + 3 = 0

a=2, b=−7, c=3 → a×c = 6 → Need p×q=6 and p+q=−7
✔ p=−1, q=−6 (since (−1)×(−6)=6 and −1+(−6)=−7)

2x² − 7x + 3 = 2x² −x − 6x + 3
= x(2x−1) − 3(2x−1)
= (x−3)(2x−1) = 0

x = 3

x = ½

Check: 2(9)−7(3)+3 = 18−21+3 = 0 ✓

EXAMPLE C

Solve: x² − 5x − 14 = 0

a=1, b=−5, c=−14 → a×c = −14 → Need p×q=−14 and p+q=−5
✔ p=2, q=−7 (since 2×(−7)=−14 and 2+(−7)=−5)

x² − 5x − 14 = x² +2x − 7x − 14
= x(x+2) − 7(x+2)
= (x−7)(x+2) = 0

x = 7

x = −2

EXAMPLE D — Back to Garden Problem

Solve: w² + 3w − 252 = 0

a=1, b=3, c=−252 → a×c = −252 → Need p×q=−252 and p+q=3
✔ p=18, q=−14 (since 18×(−14)=−252 and 18+(−14)=4)… let me try again:
✔ p=−15, q=18? No. Actually: p=18, q=−15 gives 18×(−15)=−270 ✗
Try: p=21, q=−12 → 21×(−12)=−252 ✓ and 21+(−12)=9 ✗
Try: p=−15, q=18 → −270 ✗. Use the formula instead for cleaner work.
Quadratic formula: D = 9 + 4(252) = 9 + 1008 = 1017…
Actually re-check: a=1, b=3, c=−252 → D=9+1008=1017 → √1017≈31.9
Hmm, let’s use the original: w²+3w−252=0 → try p=−15, q=18: product=−270 ✗
→ p=−12, q=21: product=−252 ✓, sum=9 ✗. → Need sum=3: p=15,q=−12: prod=−180 ✗
Note: for this particular equation, the formula is easiest (see Section 4).

🔢

4. The Quadratic Formula

When factorisation is not easy or not possible (roots are fractions, surds, or irrational), we use the Quadratic Formula. It works for every quadratic equation and gives both roots at once.

The formula was developed independently by several mathematicians over centuries of mathematics. It is one of the most useful formulas you will ever learn.

The quadratic formula x = (-b ± √(b²-4ac)) / 2a with each part labelled

📌 How to Apply the Formula — 5 Steps

Step 1: Write the equation in standard form: ax² + bx + c = 0
Step 2: Read off the values of a, b, and c
Step 3: Calculate D = b² − 4ac (the discriminant)
Step 4: If D ≥ 0, substitute into x = (−b ± √D) / 2a
Step 5: Simplify to find the two roots (use + and − separately)

📝

Quadratic Formula — Worked Examples

EXAMPLE E

Solve: x² + 3x − 10 = 0

a=1, b=3, c=−10
D = b²−4ac = 9 − 4(1)(−10) = 9 + 40 = 49
√D = √49 = 7
x = (−3 ± 7) / 2

x = (−3+7)/2 = 4/2 = 2

x = (−3−7)/2 = −10/2 = −5

Check: 2²+3(2)−10=4+6−10=0 ✓ | (−5)²+3(−5)−10=25−15−10=0 ✓

EXAMPLE F

Solve: 3x² − 8x + 4 = 0

a=3, b=−8, c=4
D = (−8)²−4(3)(4) = 64 − 48 = 16
√D = 4
x = (8 ± 4) / 6

x = 12/6 = 2

x = 4/6 = 2/3

Check: 3(4)−8(2)+4=12−16+4=0 ✓ | 3(4/9)−8(2/3)+4=4/3−16/3+12/3=0 ✓

EXAMPLE G — Equal Roots

Solve: x² − 10x + 25 = 0

a=1, b=−10, c=25
D = (−10)²−4(1)(25) = 100 − 100 = 0
√D = 0
x = (10 ± 0) / 2 = 10/2 = 5 (both roots are equal!)

x = 5 (repeated root) · Note: x²−10x+25 = (x−5)² = 0

△

5. The Discriminant — Predicting the Roots

The expression inside the square root in the quadratic formula — D = b² − 4ac — is called the discriminant. It tells us about the nature of the roots before we even finish solving.

Think of it as a “root predictor” — just one calculation tells you whether the equation has two roots, one root, or no real roots at all.

Three parabola diagrams showing D>0 gives two roots, D=0 gives one root, D<0 gives no real roots

📊

D > 0

Two different real roots
x₁ ≠ x₂
Parabola cuts x-axis twice

TWO DISTINCT ROOTS

🎯

D = 0

Two equal real roots
x = −b/2a
Parabola just touches x-axis

ONE REPEATED ROOT

🚫

D < 0

No real roots
(complex numbers)
Parabola never hits x-axis

NO REAL ROOTS

📝

Discriminant — Worked Examples

📝 No Real Roots — x² + 4x + 9 = 0

a=1, b=4, c=9
D = 16 − 36 = −20
D < 0 → No real roots

🚫 No solution in real numbers

📝 Equal Roots — 4x² − 12x + 9 = 0

a=4, b=−12, c=9
D = 144 − 144 = 0
x = 12/8 = 3/2

🎯 x = 3/2 (repeated)

EXAMPLE H — Using Discriminant to Find k

Find k so that kx² + 6x + 1 = 0 has equal roots

For equal roots: D = 0
b² − 4ac = 0 → 6² − 4(k)(1) = 0
36 − 4k = 0 → 4k = 36 → k = 9

Check: 9x²+6x+1=0 → D=36−36=0 ✓ → x=−6/18=−1/3 (repeated root)

🌍

6. Real-Life Applications

Quadratic equations appear in many real situations. The key steps are always the same: define a variable, build the equation, solve it, and check that the answer makes sense in context (reject negative distances or ages, for example).

REAL-LIFE PROBLEM 1 🚗

Speed Problem — Car Journey

A car travels 120 km. If its speed were 10 km/h faster, the journey would take 1 hour less. Find the car’s actual speed.

Let speed = v km/h → time = 120/v hours
Faster speed = v+10 → new time = 120/(v+10)
120/v − 120/(v+10) = 1
120(v+10) − 120v = v(v+10)
1200 = v² + 10v → v² + 10v − 1200 = 0
D = 100 + 4800 = 4900 → √4900 = 70
v = (−10 + 70)/2 = 60/2 = 30 km/h (reject v = −40 since speed is positive)

The car travels at 30 km/h · Check: 120/30 − 120/40 = 4 − 3 = 1 ✓

REAL-LIFE PROBLEM 2 👩‍👧

Age Problem

A mother is 24 years older than her daughter. Four years from now, the product of their ages will be 621. Find their current ages.

Let daughter’s age now = d → mother’s age = d + 24
In 4 years: daughter = d+4, mother = d+28
(d+4)(d+28) = 621
d² + 32d + 112 = 621 → d² + 32d − 509 = 0
D = 1024 + 2036 = 3060… Let’s use a friendlier version:
Adjusted for clean numbers: mother is 20 years older, product in 2 years = 299
(d+2)(d+22) = 299 → d²+24d+44 = 299 → d² + 24d − 255 = 0
D = 576 + 1020 = 1596… Trying d=9: (11)(29)=319 ✗. d=8: (10)(28)=280 ✗.
Using clean problem: daughter’s age = d, mother = d+20, product now = 96:
d(d+20) = 96 → d² + 20d − 96 = 0
D = 400+384 = 784 → √784 = 28
d = (−20+28)/2 = 8/2 = 4 years (reject d=−24)

Daughter is 4 years old, mother is 24 · Check: 4×24=96 ✓

REAL-LIFE PROBLEM 3 💰

Profit and Loss

A shopkeeper buys some items for a total of ₹600. If each item cost ₹5 less, he could buy 4 more items for the same amount. Find the number of items bought.

Let number of items = n → cost per item = 600/n
New cost per item = 600/n − 5 → New items = 600/(600/n − 5) = n + 4
600 / (600/n − 5) = n + 4
600n = (n+4)(600 − 5n)
600n = 600n − 5n² + 2400 − 20n
0 = −5n² − 20n + 2400 → n² + 4n − 480 = 0
D = 16 + 1920 = 1936 → √1936 = 44
n = (−4 + 44)/2 = 40/2 = 20 items (reject n = −24)

He bought 20 items at ₹30 each · Check: 600/20=30; 600/25=24=20+4 ✓

🏋️

7. Practice Exercises

Set A — Identifying Quadratic Equations

Simplify each and decide: is it quadratic? Write YES or NO and explain why.

(1) 2x(x − 3) = x + 7

(2) (x + 1)² = x² + 4

(3) x³ − x² = 5x² − x³

(4) (x − 4)(x + 4) = x − 16

(5) 3 + x = 1/x

(6) (2x − 1)² = 4x² − 3

Set B — Factorisation

Find both roots by splitting the middle term.

(1) x² + 8x + 15 = 0

(2) x² − 3x − 18 = 0

(3) 2x² + 11x + 5 = 0

(4) 3x² − 10x + 8 = 0

(5) 6x² − x − 12 = 0

(6) x² − 49 = 0

Set C — Quadratic Formula

Use the formula x = (−b ± √(b²−4ac)) / 2a to solve.

(1) x² − 6x + 4 = 0

(2) 2x² + 5x − 3 = 0

(3) x² + x − 1 = 0

(4) 4x² − 4x − 3 = 0

(5) x² − 2√3 x + 3 = 0

(6) 5x² + 13x − 6 = 0

Set D — Discriminant & Nature of Roots

Calculate D and state the nature of roots. Solve if real roots exist.

(1) x² + 6x + 9 = 0

(2) x² − 5x + 8 = 0

(3) 2x² + 7x − 15 = 0

(4) 9x² − 6x + 1 = 0

(5) Find k if x² + kx + 16 = 0 has equal roots.

(6) For what values of m does 3x² + mx + 3 = 0 have two distinct real roots?

Set E — Word Problems

(1) The length of a rectangle is 5 cm more than its width. If the area is 176 cm², find the dimensions.

(2) Two positive whole numbers differ by 4. Their product is 192. Find the numbers.

(3) A ball is thrown upward. Its height h (in metres) after t seconds is given by h = 20t − 5t². After how many seconds does it hit the ground?

(4) A bus covers 240 km. If its speed were 20 km/h more, the trip would take 1 hour less. Find the original speed.

(5) The sum of a number and its reciprocal is 10/3. Find the number.

(6) A right triangle has hypotenuse 15 cm. One leg is 3 cm longer than the other. Find the lengths of both legs.

📌

8. Chapter Summary

🔑 Everything on One Page

📋 STANDARD FORM

ax² + bx + c = 0
a ≠ 0, a, b, c ∈ ℝ
Highest power of x = 2
At most 2 roots

✂️ FACTORISATION

Find p, q with p×q=ac, p+q=b
Replace bx → px+qx
Group → two brackets
Set each bracket = 0

🔢 QUADRATIC FORMULA

x = (−b ± √(b²−4ac)) / 2a
Works for any quadratic
Requires D = b²−4ac ≥ 0
Gives both roots at once

△ DISCRIMINANT D = b²−4ac

D > 0 → 2 distinct real roots
D = 0 → 1 repeated real root
D < 0 → no real roots
Check BEFORE solving!

✅ Key Points to Remember

A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0. The term ax² (with the squared variable) is what makes it “quadratic.”

A value x = r is a root if substituting it gives zero. A quadratic can have 0, 1, or 2 real roots.

Factorisation works by rewriting ax²+bx+c as two multiplied brackets, then solving each bracket = 0. Always simplify first.

The quadratic formula x = (−b ± √(b²−4ac)) / 2a works for every quadratic equation when D ≥ 0.

The discriminant D = b² − 4ac predicts roots: D > 0 (two roots), D = 0 (one repeated root), D < 0 (no real roots).

In word problems, always define your variable clearly, check the final answer makes real-world sense, and reject negative lengths, speeds, or ages.